Pattab

Answered

2022-07-16

How could I solve this chain rule problem?

To find: $\frac{d}{dt}[f(\mathbf{\text{c}}(t))]{|}_{t=0}$

Where $\mathbf{\text{c}}(t)$ is such that $\frac{d}{dt}\mathbf{\text{c}}(t)=\mathbf{\text{F}}(c(t))$, with $\mathbf{\text{F}}=\mathrm{\nabla}f$, $\mathbf{\text{c}}(0)=(-\frac{\pi}{2},\frac{\pi}{2})$ and $f(x,y)=\mathrm{sin}(x+y)xy$

I think I would need to proceed using the chain rule but I am currently not sure. How would I go about solving this problem?

Thanks in advance

To find: $\frac{d}{dt}[f(\mathbf{\text{c}}(t))]{|}_{t=0}$

Where $\mathbf{\text{c}}(t)$ is such that $\frac{d}{dt}\mathbf{\text{c}}(t)=\mathbf{\text{F}}(c(t))$, with $\mathbf{\text{F}}=\mathrm{\nabla}f$, $\mathbf{\text{c}}(0)=(-\frac{\pi}{2},\frac{\pi}{2})$ and $f(x,y)=\mathrm{sin}(x+y)xy$

I think I would need to proceed using the chain rule but I am currently not sure. How would I go about solving this problem?

Thanks in advance

Answer & Explanation

iskakanjulc

Expert

2022-07-17Added 18 answers

Let $g=f\circ c$, i.e. let

$g(t)=f(c(t))=f({c}_{1}(t),{c}_{2}(t)).$

Then by the chain rule

${g}^{\prime}(t)={f}_{x}(c(t)){c}_{1}^{\prime}(t)+{f}_{y}(c(t)){c}_{2}^{\prime}(t)=\mathrm{\nabla}f(c(t))\cdot {c}^{\prime}(t)=\mathrm{\nabla}f(c(t))\cdot \mathrm{\nabla}f(c(t)),$

where the second equality follows from the assumption about ${c}^{\prime}(t)$

Since

${f}_{x}(x,y)=y[\mathrm{sin}(x+y)+x\mathrm{cos}(x+y)]$

and

${f}_{y}(x,y)=x[\mathrm{sin}(x+y)+y\mathrm{cos}(x+y)]$

we have

${f}_{x}(c(0))={f}_{x}(-\frac{\pi}{2},\frac{\pi}{2})=-\frac{{\pi}^{2}}{4}$

${f}_{y}(c(0))={f}_{y}(-\frac{\pi}{2},\frac{\pi}{2})=-\frac{{\pi}^{2}}{4}$

or

$\mathrm{\nabla}f(c(0))=(-\frac{{\pi}^{2}}{4},-\frac{{\pi}^{2}}{4}).$

Thus

${g}^{\prime}(0)=\mathrm{\nabla}f(c(0))\cdot \mathrm{\nabla}f(c(0))=(-\frac{{\pi}^{2}}{4},-\frac{{\pi}^{2}}{4})\cdot (-\frac{{\pi}^{2}}{4},-\frac{{\pi}^{2}}{4})=\frac{{\pi}^{4}}{8}.$

$g(t)=f(c(t))=f({c}_{1}(t),{c}_{2}(t)).$

Then by the chain rule

${g}^{\prime}(t)={f}_{x}(c(t)){c}_{1}^{\prime}(t)+{f}_{y}(c(t)){c}_{2}^{\prime}(t)=\mathrm{\nabla}f(c(t))\cdot {c}^{\prime}(t)=\mathrm{\nabla}f(c(t))\cdot \mathrm{\nabla}f(c(t)),$

where the second equality follows from the assumption about ${c}^{\prime}(t)$

Since

${f}_{x}(x,y)=y[\mathrm{sin}(x+y)+x\mathrm{cos}(x+y)]$

and

${f}_{y}(x,y)=x[\mathrm{sin}(x+y)+y\mathrm{cos}(x+y)]$

we have

${f}_{x}(c(0))={f}_{x}(-\frac{\pi}{2},\frac{\pi}{2})=-\frac{{\pi}^{2}}{4}$

${f}_{y}(c(0))={f}_{y}(-\frac{\pi}{2},\frac{\pi}{2})=-\frac{{\pi}^{2}}{4}$

or

$\mathrm{\nabla}f(c(0))=(-\frac{{\pi}^{2}}{4},-\frac{{\pi}^{2}}{4}).$

Thus

${g}^{\prime}(0)=\mathrm{\nabla}f(c(0))\cdot \mathrm{\nabla}f(c(0))=(-\frac{{\pi}^{2}}{4},-\frac{{\pi}^{2}}{4})\cdot (-\frac{{\pi}^{2}}{4},-\frac{{\pi}^{2}}{4})=\frac{{\pi}^{4}}{8}.$

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